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Mechanics

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20 Nov 2007 19:52:20 IST

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2962

IIT 96 sum (Newtons)

None

A Smooth semi-circular wire track of radius R is fixed in a
vertical plane.One end of a massless spring of natural length
(3R/4) is attached to the lower point of O of the wire track

A small ring of mass 'm',which can slide on the track , is
attached to the other end of the spring.The ring is held
stationary at P such that spring akes an angle 60⁰ with the vertical.
The spring constant K = mg/R.

Consider the instant when the ring is released.

a)Determine the tangential acceleration of the ring and the
normal reaction.

The one who answers gets 2 salutes from me.
Challenge for all !!!!!!!!!!

(If answered then salutes but if no answer then I will give
the solution by tomorrow)

Cheers!!!!!!@@@!!!!!!!!!!!!!

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waterdemon

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20 Nov 2007 20:23:59 IST

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Is nobody able to do this......!!!!!!!!!!!

Cmon where are the geniouses who were discussing Rotation
yesterday.??

Rohan

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20 Nov 2007 20:29:06 IST

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what's the answer ?? ??

Karthik M

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20 Nov 2007 20:29:55 IST

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Wait... just saw the sum... trying my shot at it... dont give the answer now

Tarun Sharma

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20 Nov 2007 20:31:24 IST

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normal reaction is mg(8cos

- 1 ) / 8

Rohan

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20 Nov 2007 20:31:42 IST

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don't give the solution

but just tell the answer

u can also send it to my nudgebook if u don't want it to be here....

waterdemon

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20 Nov 2007 20:31:47 IST

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The answers are:
N=????
Tangential acceleration = ????

Try solving..............will post the solution by tomorrow.
Cheers!!!!!!!!!!!!!!!

As Karthik said not to answer.

Karthik M

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20 Nov 2007 20:35:23 IST

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Is the point P Touching the line passing thru C?

Niteesh Mehra

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20 Nov 2007 20:48:53 IST

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this question is useless yaar..aint it just too tuf ??

Rahul Sharma

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20 Nov 2007 20:50:26 IST

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Will it be like,the initial tangential acc will be downwards, assuming that point P to be on the horizontal line from the center?

So m(aT) = mg + kx cos60 ?

plz correct me if i am wrong.

Tarun Sharma

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20 Nov 2007 20:55:36 IST

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no nitish,,,,

Rahul Sharma

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20 Nov 2007 20:57:30 IST

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hey, plz tell me if i am doing alrite.

Karthik M

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20 Nov 2007 20:59:30 IST

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Is alpha = g(8-

3/4

3)/R???

nishant singh

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20 Nov 2007 22:19:05 IST

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normal reaction = mg(1/2 - sqrt(3)/8)

tangential acceleration = g(sqrt(3)/2 + 1/8)

centripetal acceleration = g(1/2 - sqrt(3)/8)
am i rite????

Rahul Sharma

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20 Nov 2007 22:45:13 IST

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Is the tangential acceleration 25root3/4 ?

a b

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20 Nov 2007 22:59:28 IST

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Dont give answer till tomorrow 6 pm. Got exam at noon. Will try after that.So, waterdemon, dont give answer bfor that please.

waterdemon

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20 Nov 2007 23:35:46 IST

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OK!!!!!!!!!!
Good night !!!!!!!
Answer after 6.00pm

Aniket

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21 Nov 2007 11:35:21 IST

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normal reaction=3mg/8 and tangential accn.=5

3g/8

Rahul Sharma

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21 Nov 2007 11:52:50 IST

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Yup my answers are also the same as frenied's.

aT = 5root3 g/8 or 25root3/4 and N = 3mg/8 or 15/4 m.

karthik 007

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21 Nov 2007 12:12:07 IST

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hey the ans is:
N=3mg/8
a(tangential)=5*1.73*g/8
this is how i solved

i resolved all the forces along the centre and tangent
from this i got
N+kxcos60=mgcos60
along tangent
ma=(kx+mg)sin60

now calculate x
x=r/4(hint: equilateral triangle is formed)
solve and get the answer

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